Unit 3: Real Sequences: Limits, Monotone Convergence & Cauchy Sequences

1. Real Sequences as Functions

Definition 3.1 (Real Sequence): A sequence of real numbers is a function $a: \mathbb{N} \to \mathbb{R}$. We typically denote the values $a(n)$ as $a_n$ (the $n$-th term), and the sequence itself by $\{a_n\}_{n=1}^\infty$, $(a_n)_{n \in \mathbb{N}}$, or simply $(a_n)$.

2. The $\epsilon$-$N$ Definition of Sequence Convergence

The concept of a limit is the foundational bedrock of all mathematical analysis, formalized by Augustin-Louis Cauchy and Karl Weierstrass to eliminate vague notions of "infinitesimal approaches."

Definition 3.2 (Limit of a Sequence / Convergence): A sequence $(a_n)$ is said to converge to a real number $L \in \mathbb{R}$, denoted:

if for every $\epsilon > 0$, there exists an index $N \in \mathbb{N}$ (dependent on $\epsilon$) such that:

In topological terms, for every $\epsilon$-neighborhood $V_\epsilon(L)$, there exists $N \in \mathbb{N}$ such that $a_n \in V_\epsilon(L)$ for all $n \ge N$. If no such $L \in \mathbb{R}$ exists, the sequence is said to diverge.

3. Theorem: Uniqueness of Sequence Limits

Can a sequence converge to two distinct real numbers simultaneously? In any Hausdorff space—and specifically $\mathbb{R}$—the answer is strictly no.

Theorem 3.1 (Uniqueness of Limits): If a sequence $(a_n)$ converges, its limit is unique. That is, if $a_n \to L_1$ and $a_n \to L_2$, then $L_1 = L_2$.

Line-by-Line Proof:

We prove this by contradiction. Suppose $a_n \to L_1$ and $a_n \to L_2$ with $L_1 \ne L_2$. Then $|L_1 - L_2| > 0$. Choose:

Since $a_n \to L_1$, there exists $N_1 \in \mathbb{N}$ such that:

Since $a_n \to L_2$, there exists $N_2 \in \mathbb{N}$ such that:

Let $N = \max\{N_1, N_2\}$. For any $n \ge N$, both inequalities hold simultaneously. Using the Triangle Inequality:

Substituting the $\epsilon$-bounds:

This yields the strict inequality $|L_1 - L_2| < |L_1 - L_2|$, an impossible contradiction! Therefore, $L_1 = L_2$. $\blacksquare$

4. Convergence Implies Boundedness

Theorem 3.2 (Boundedness of Convergent Sequences): Every convergent sequence $(a_n)$ is bounded.

Proof:

Suppose $a_n \to L$. Setting $\epsilon = 1 > 0$, by the definition of limit there exists $N \in \mathbb{N}$ such that:

For indices prior to $N$ (the terms $a_1, a_2, \dots, a_{N-1}$), there are only finitely many values. Define:

Then for all $n \in \mathbb{N}$, $|a_n| \le M$. Thus, $(a_n)$ is bounded. $\blacksquare$ (Note: The converse is false! The sequence $a_n = (-1)^n$ is bounded by 1, but diverges).

1. Algebraic Limit Theorems

Theorem 3.3 (Algebra of Limits): Let $(a_n)$ and $(b_n)$ be convergent sequences with $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$.

1. Sum Rule: $\lim_{n \to \infty} (a_n + b_n) = A + B$.
2. Scalar Multiple: $\lim_{n \to \infty} (c a_n) = c A$ for any constant $c \in \mathbb{R}$.
3. Product Rule: $\lim_{n \to \infty} (a_n b_n) = A B$.
4. Quotient Rule: If $B \ne 0$ and $b_n \ne 0$ for all $n$, then $\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right) = \frac{A}{B}$.

Rigorous Proof of the Product Rule:

We rewrite the difference $a_n b_n - AB$ by adding and subtracting an intermediate term $a_n B$:

Applying the Triangle Inequality:

By Theorem 3.2, since $(a_n)$ converges, it is bounded: there exists $M > 0$ such that $|a_n| \le M$ for all $n \in \mathbb{N}$. Thus:

Let $\epsilon > 0$ be given. Since $a_n \to A$, choose $N_1 \in \mathbb{N}$ such that $|a_n - A| < \frac{\epsilon}{2(|B| + 1)}$ for all $n \ge N_1$. Since $b_n \to B$, choose $N_2 \in \mathbb{N}$ such that $|b_n - B| < \frac{\epsilon}{2M}$ for all $n \ge N_2$. Set $N = \max\{N_1, N_2\}$. For all $n \ge N$:

This proves that $a_n b_n \to AB$. $\blacksquare$

2. Order Properties and the Squeeze Theorem

Theorem 3.4 (Order Limit Theorem): If $a_n \to A$ and $b_n \to B$, and there exists $N_0 \in \mathbb{N}$ such that $a_n \le b_n$ for all $n \ge N_0$, then:

Theorem 3.5 (The Squeeze / Sandwich Theorem): Let $(a_n)$, $(b_n)$, and $(c_n)$ be sequences such that:

If $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} c_n = L$, then:

Proof of Squeeze Theorem:

Let $\epsilon > 0$. Since $a_n \to L$, there exists $N_1$ such that $|a_n - L| < \epsilon \iff L - \epsilon < a_n < L + \epsilon$ for all $n \ge N_1$. Since $c_n \to L$, there exists $N_2$ such that $|c_n - L| < \epsilon \iff L - \epsilon < c_n < L + \epsilon$ for all $n \ge N_2$. Choose $N = \max\{N_0, N_1, N_2\}$. For all $n \ge N$:

Thus $b_n \to L$. $\blacksquare$

1. Monotone Sequences

Definition 3.3 (Monotonicity): A sequence of real numbers $(a_n)$ is:

1. Monotonically increasing (or non-decreasing) if $a_{n} \le a_{n+1}$ for all $n \in \mathbb{N}$.
2. Strictly increasing if $a_n < a_{n+1}$ for all $n \in \mathbb{N}$.
3. Monotonically decreasing (or non-increasing) if $a_n \ge a_{n+1}$ for all $n \in \mathbb{N}$.
4. Strictly decreasing if $a_n > a_{n+1}$ for all $n \in \mathbb{N}$.

A sequence is monotone if it is either monotonically increasing or monotonically decreasing.

2. The Monotone Convergence Theorem

The Monotone Convergence Theorem is one of the most powerful analytical engines in real analysis, enabling us to prove convergence without knowing the numerical value of the limit in advance.

Theorem 3.6 (Monotone Convergence Theorem - MCT): A monotone sequence of real numbers converges if and only if it is bounded:

1. If $(a_n)$ is monotonically increasing and bounded above, then:

1. If $(a_n)$ is monotonically decreasing and bounded below, then:

Line-by-Line Proof:

$(\implies)$ By Theorem 3.2, every convergent sequence is bounded. $(\impliedby)$ Assume $(a_n)$ is monotonically increasing and bounded above. Define the non-empty set of values:

Since $(a_n)$ is bounded above, $S$ is bounded above. By the Completeness Axiom of $\mathbb{R}$, $S$ has a supremum:

We now prove that $\lim_{n \to \infty} a_n = L$. Let $\epsilon > 0$ be given. Since $L$ is the least upper bound of $S$, $L - \epsilon$ is not an upper bound of $S$. Therefore, there exists some element $a_N \in S$ such that:

Since $(a_n)$ is monotonically increasing, for all $n \ge N$:

Furthermore, since $L$ is an upper bound of $S$, for all $n \in \mathbb{N}$:

Combining both inequalities, for all $n \ge N$:

This precisely satisfies the $\epsilon$-$N$ definition of limit. Therefore, $\lim_{n \to \infty} a_n = L = \sup \{a_n\}$. The proof for a decreasing bounded sequence follows symmetrically by considering the infimum. $\blacksquare$

1. Subsequences

Definition 3.4 (Subsequence): Let $(a_n)$ be a sequence. Let $(n_k)_{k=1}^\infty$ be a strictly increasing sequence of natural numbers:

The sequence $(a_{n_k})_{k=1}^\infty$ is called a subsequence of $(a_n)$. Notice that $n_k \ge k$ for all $k \in \mathbb{N}$.

Theorem 3.7 (Convergence of Subsequences): If a sequence $(a_n)$ converges to $L$, then every subsequence $(a_{n_k})$ also converges to $L$.

2. The Bolzano-Weierstrass Theorem for Sequences

Theorem 3.8 (Bolzano-Weierstrass Theorem for Sequences): Every bounded sequence of real numbers has a convergent subsequence.

Rigorous Proof:

We present the classical proof using the Peak Point Lemma (Monotone Subsequence Theorem):

Lemma (Peak Point Lemma): Every sequence $(a_n)$ contains a monotone subsequence. Proof of Lemma: We say that an index $m \in \mathbb{N}$ is a peak (or turn point) of the sequence $(a_n)$ if:

That is, $a_m$ is greater than or equal to all subsequent terms. Now consider two mutually exclusive cases:

• Case 1: There are infinitely many peaks.

Let the peaks be indexed as $m_1 < m_2 < m_3 < \dots < m_k < \dots$. By definition of a peak, since $m_{k+1} > m_k$, we have $a_{m_k} \ge a_{m_{k+1}}$. Thus, the subsequence $(a_{m_k})_{k=1}^\infty$ is monotonically decreasing!

• Case 2: There are only finitely many peaks (or no peaks at all).

Then there exists an index $N_0 \in \mathbb{N}$ after which no peaks exist. Choose $n_1 > N_0$. Since $n_1$ is not a peak, there exists an index $n_2 > n_1$ such that $a_{n_1} < a_{n_2}$. Similarly, since $n_2 > N_0$, $n_2$ is not a peak, so there exists $n_3 > n_2$ such that $a_{n_2} < a_{n_3}$. Continuing inductively, we produce a strictly increasing sequence of indices $n_1 < n_2 < n_3 < \dots$ such that:

Thus, the subsequence $(a_{n_k})_{k=1}^\infty$ is strictly increasing! In either case, $(a_n)$ contains a monotone subsequence $(a_{n_k})$. $\blacksquare$

Now, to complete the Bolzano-Weierstrass proof: Let $(a_n)$ be a bounded sequence. By the Peak Point Lemma, $(a_n)$ contains a monotone subsequence $(a_{n_k})$. Since $(a_n)$ is bounded, the subsequence $(a_{n_k})$ is also bounded. By the Monotone Convergence Theorem (Theorem 3.6), this bounded monotone subsequence must converge! Therefore, $(a_n)$ has a convergent subsequence. $\blacksquare$

3. Limit Superior and Limit Inferior

For sequences that oscillate and do not converge (such as $a_n = (-1)^n$), the ordinary limit does not exist. However, every bounded sequence possesses a well-defined $\limsup$ and $\liminf$.

Definition 3.5 ($\limsup$ and $\liminf$): Let $(a_n)$ be a bounded sequence of real numbers. For each $k \in \mathbb{N}$, define the tail bounds:

Notice that as $k$ increases, the set $\{a_n : n \ge k\}$ shrinks, so $(u_k)$ is monotonically decreasing and bounded below, while $(v_k)$ is monotonically increasing and bounded above. By MCT, both limits exist. We define:

Theorem 3.9 (Convergence Criterion via Limsup and Liminf): A bounded sequence $(a_n)$ converges if and only if:

in which case $\lim_{n \to \infty} a_n = L$.

1. Cauchy Sequences

In many applications, evaluating the limit $L$ of a sequence directly is impossible because $L$ is unknown. Augustin-Louis Cauchy introduced a revolutionary criterion: instead of testing how close terms get to an external limit $L$, test how close the terms get to each other!

Definition 3.6 (Cauchy Sequence): A sequence $(a_n)$ is a Cauchy sequence if for every $\epsilon > 0$, there exists an index $N \in \mathbb{N}$ such that:

2. Properties of Cauchy Sequences

Lemma 3.1 (Convergent Sequences are Cauchy): If $\lim_{n \to \infty} a_n = L$, then $(a_n)$ is a Cauchy sequence.

Proof: Let $\epsilon > 0$. Since $a_n \to L$, choose $N$ such that $|a_k - L| < \frac{\epsilon}{2}$ for all $k \ge N$. For any $n, m \ge N$:

Lemma 3.2 (Cauchy Sequences are Bounded): Every Cauchy sequence is bounded.

Proof: Let $\epsilon = 1$. Choose $N$ such that $|a_n - a_m| < 1$ for all $n, m \ge N$. In particular, setting $m = N$: $|a_n - a_N| < 1 \implies |a_n| < |a_N| + 1$ for all $n \ge N$. Taking $M = \max\{|a_1|, |a_2|, \dots, |a_{N-1}|, |a_N| + 1\}$ bounds the sequence for all $n$. $\blacksquare$

3. The Cauchy Completeness Criterion

Theorem 3.10 (Cauchy Completeness of $\mathbb{R}$): A sequence of real numbers $(a_n)$ converges in $\mathbb{R}$ if and only if it is a Cauchy sequence.

Complete Rigorous Proof:

$(\implies)$ Proved in Lemma 3.1. $(\impliedby)$ Let $(a_n)$ be a Cauchy sequence in $\mathbb{R}$.

1. Find a candidate limit $L$:

By Lemma 3.2, $(a_n)$ is bounded. By the Bolzano-Weierstrass Theorem (Theorem 3.8), $(a_n)$ contains a convergent subsequence $(a_{n_k})$. Let:

2. Prove that the entire sequence $(a_n)$ converges to $L$:

Let $\epsilon > 0$ be arbitrary. Since $(a_n)$ is Cauchy, choose $N_1 \in \mathbb{N}$ such that:

Since $a_{n_k} \to L$, choose $K \in \mathbb{N}$ such that:

Now pick an index $k \ge K$ large enough so that $n_k \ge N_1$ (which is always possible since $n_k \ge k$). Then for any index $n \ge N_1$, we set $m = n_k$:

Since $n \ge N_1$ and $n_k \ge N_1$, $|a_n - a_{n_k}| < \frac{\epsilon}{2}$. And since $k \ge K$, $|a_{n_k} - L| < \frac{\epsilon}{2}$. Therefore:

Since this holds for all $n \ge N_1$, we conclude that:

Thus, every Cauchy sequence in $\mathbb{R}$ converges. $\blacksquare$